Two functions are given. f(x)=-x^2+3x-2

g(x)=-x+1

What is one value for x such that f(x)=g(x)?

Skip to content# Two functions are given.f(x)=-x^2+3x-2g(x)=-x+1What is one value for x such that f(x)=g(x)?

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Two functions are given. f(x)=-x^2+3x-2

g(x)=-x+1

What is one value for x such that f(x)=g(x)?

Hi

f(x) = g(x) if -x²+3x-2 - ( -x+1) = 0

-x² +3x-2 +x-1 = 0

-x² +4x -3 = 0

To solve, tou have to use the general method of resolution of a quadratic fonction.

To determine if it's has a solution in R, let's calculate Δ

Δ = (4)² - 4 * (1) *(-3)

Δ = 16 +12

Δ= 28

as Δ≥ 0 so the function allow two solution within R

so S 1 = ( -4 +√28) / 2 S 2 = (-4 -√28 ) /2

S1 = ( -4 + 2√7) /2 S2 = (-4 - 2√7) /2

S1 = (2 (-2 +√7) /2 S2 2 (-2 -√7) /2

S1 = -2 +√7 S2 = -2 -√7

So the two function are equal twice. one for x = -2 +√7 and second x = -2-√7

Iwould say rat and vsu but i’m not 100% sure

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice